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OpenStudy (goten77):
kinda obvious what x is but i dont kno how to explain or do this 1 mathmetically
hartnn (hartnn):
if |y|=a
then \[y=\pm a\]
so |x+1|=2x
gives (x+1)= 2x or (x+1)=-2x
can u solve these two to get 2 values of x ??
out of these 2 ,one after re substituting in original equation,will not give u equality.so u discard that vale.
then u will have only 1 solution.
OpenStudy (anonymous):
@hartnn so how do i do |x+1|=2x
OpenStudy (anonymous):
subtract 1 from both sides? @hartnn
hartnn (hartnn):
as i have explained,u have
(x+1)= 2x or (x+1)=-2x
u need to solve both of these
take one at a time
x+1=2x
subtract x from both sides,what u get??
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OpenStudy (anonymous):
1x? @hartnn
OpenStudy (anonymous):
right?
hartnn (hartnn):
1x where on left side or right side?
OpenStudy (anonymous):
right side @hartnn
hartnn (hartnn):
so whats on left side??
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OpenStudy (anonymous):
um is it 1
hartnn (hartnn):
yup,so u have
1x=1
that is
x=1
is one of your solution.
can u get other solution similarly ?
OpenStudy (anonymous):
Thanks and i'll try
OpenStudy (anonymous):
so the second on is |x+1|=-2x right? @hartnn
hartnn (hartnn):
(x+1)=-2x
u can remove mod | | sign,since u are taking -2x
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OpenStudy (anonymous):
is x=-3 correct?
OpenStudy (anonymous):
@hartnn
hartnn (hartnn):
nopes,
u get 3x=-1
so x=-(1/3)
but when u put this in original equation,equality does not get satisfied,
so the only solution u have is
x=1
ok?